Question

40.0 mL of 0.200 N NaOH is mixed with 60.0 mL of 0.300 N HCl. What is the normality of the remaining H+ if the final volume is 100.0 mL?

Answer 1

Total number of gram equivalents (geqv) of NaOH added = Volume X Normality = 40.0mL X 0.200N = 8.00 mgeqv = 0.008 mgeqv

similarly

Total number of gram equivalents (geqv) of HCl = 60mL X 0.300N = 18.00 mgeqv = 0.018 geqv

out of 0.018 gram equivalents of HCl 0.008 will react be neutralised by NaOH.

Therefore gram equivalents of HCl remained = 0.018 - 0.008 = 0.01 Gram equivalent

Normality (N) of HCl = Number of gram equivalents of HCl remained unreacted / Volume in liters

Volume = 40mL + 60mL = 100mL = 0.1L

Hence N = 0.01 / 0.1 = 0.1N